Alright, so from what I gather, most of the first part was ok besides the fact that is the upper bound and not .

As for the second part...
Let and . Then,
since is an upper bound for and f is increasing then there is no such that . Thus, . If we let this limit be called then we have that where . Hence, we have that .

Sorry, too many flaws. The second part of the problem -- proving -- has nothing to do with f(a). In the example I gave above, f(a) is different from .

Let and .

Are you assuming? In this case, since this assumption is not in the problem statement, you later need to consider the possibility that . However, I thought that you have proved this in part 1. Therefore, you should say, "Let . We have already shown that ." (To repeat, the second sentence is irrelevant.)

Then, since is an upper bound for and f is increasing then there is no such that .

What happens to the right of is irrelevant to the problem.

Thus, .

I don't see how this follows from the previous discussion.

If we let this limit be called

This is redefining A.

then we have that where . Hence, we have that .

It looks like this conclusion is based on confusing two meanings of A.

Let us define , as you have done. Here is how I would prove that . We show that for every there exists a such that when and we have . Note that the second inequality, i.e., , holds by the definition of .

Let be given. Also by the definition of , is not an upper bound of ; therefore, there exists an such that . But then, since f is increasing, for every such that we have . Therefore, we can take .

A quasi-continuous function is one for which the right and left hand limits exist at each point. It is a standard theorem in analysis that any monotonic function is quasi-continuous. That is the essence of this problem. You are asked to show that the limit on the left exist at each point.